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drkoverlord
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Help with Stoich limiting reactant Lab urgent!
1. I need to find the mass of the excess reactant. This was a lab i did. so its BaCl2.2H20 + Na2So4 --> BaSO4 + 2NaCL. The data i have calculated thus far is.
moles of BaSO4=0.0019 mol
moles of BaCl2=0.0019 mol
moles of Na2SO4=0.0019 mol
mass of BaSo4=.4524g
mass of BaCl2=.4641g
mass of Na2So4= .2698g
Mass of entire salt mixture=1.0038g
The limiting reactant is SO4 and excess is Ba.
BaSo4 is the precipitate.
How do i get the excess reactant!
1. I need to find the mass of the excess reactant. This was a lab i did. so its BaCl2.2H20 + Na2So4 --> BaSO4 + 2NaCL. The data i have calculated thus far is.
moles of BaSO4=0.0019 mol
moles of BaCl2=0.0019 mol
moles of Na2SO4=0.0019 mol
mass of BaSo4=.4524g
mass of BaCl2=.4641g
mass of Na2So4= .2698g
Mass of entire salt mixture=1.0038g
The limiting reactant is SO4 and excess is Ba.
BaSo4 is the precipitate.
How do i get the excess reactant!